Number Theory and Cryptography

CMPS/MATH 2170: Discrete Mathematics

Outline

• Divisibility and Modular Arithmetic (4.1)• Primes and GCD (4.3)• Solving Congruences (4.4)• Cryptography (4.6)

Division

Definition: Let !, # ∈ ℤ with ! ≠ 0. we say ! divides # if #/! ∈ ℤ• equivalently, # = *! for some * ∈ ℤ• we use ! | # to denote ! divides # (or # is divisible by !) • if ! | #, we say that ! is a factor or divisor of #

Ex. 1: Determine whether

a. 3 | 7

b. 3 | 12

Ex. 2: How many positive integers not exceeding , are divisible by 3? ,/3

Division (cont.)

Theorem: Let !, #, $ ∈ ℤ and ! ≠ 0. Then

(i) If ! | # and ! | $, then ! | (# + $)(ii) If ! | #, then ! | #$(iii) If ! | # and # | $ (# ≠ 0) , then ! | $

Prime Numbers

Definition: An integer ! > 1 is called prime if the only positive factors of !are 1 and !• ! is prime ⇔ ∀& ∈ ℤ): & | ! → & = 1 or & = !

Definition: An integer > 1 that is not prime is called composite

• 1 is neither prime nor composite

The Fundamental Theorem of Arithmetic

Theorem: Every positive integer > 1 can be written uniquely as a prime or as the product of two or more primes written in a non-decreasing order • “prime factorization of an integer”

Ex: 100 =641 =999 =

Proof of the fundamental theorem:

1. existence: strong induction

2. uniqueness: to be proved

q prime factorization is hard for large numbers2 ⋅ 2 ⋅ 5 ⋅ 5 = 2* ⋅ 5*

6413 ⋅ 3 ⋅ 3 ⋅ 37 = 3- ⋅ 37

Applications of the Fundamental Theorem

Theorem: A composite ! has a prime divisor ≤ !.

Corollary: An integer # > 1 is a prime if it is not divisible by any prime ≤ #.

Ex: Show that 101 is prime

Theorem: There are infinitely many primes

• A proof given by Euclid in The Elements

Two Great Open Problems on Primes

• Goldbach’s conjecture (1742): every even number ! > 2 is the sum of two primes• Every even number ! > 2 is the sum of at most 6 primes (1995)

• Every even number ! > 2 is the sum of a prime and a number that is either prime or the product of two primes (1+2, 1966)

• Twin prime conjecture (before 1849): there are infinitely many twin primes• Twin prime pairs: (3, 5), (5,7), (11, 13), (17, 19), (29, 31), …

• There are infinitely many pairs of prime numbers that differ by 246 or less (2014)

Greatest Common Divisors

Definition: Let !, # ∈ ℤ, not both zero. The largest integer & such that & | ! and & | # is called the greatest common divisor of ! and #, denoted by & = gcd(!, #)Ex: gcd 24, 36 =

gcd 17, 22 =gcd 120, 500 =

gcd 6789 ⋅ 6;8< ⋯6>8?, 67@9 ⋅ 6;@< ⋯6>@? = 67ABC(89,@9) ⋅ 6;ABC(8<,@<) ⋯6>ABC(8?,@?)

• Is there a more efficient way to find gcd?

121gcd(2D ⋅ 3 ⋅ 5, 2; ⋅ 5D) = 2; ⋅ 5 = 20

Least Common Multiples

Let !, # ∈ ℤ, !, # ≠ 0. The smallest positive integer that is divisible by both ! and # is called the least common multiple of ! and #, denoted by lcm(!, #)

Ex: lcm 24, 36 =

lcm(2345 ⋅ 2748 ⋯2:4;, 23<5 ⋅ 27<8 ⋯2:<;) = 23=>?(45,<5) ⋅ 27=>? (48,<8) ⋯2:=>?(4;,<;)

Theorem: For any positive integers ! and #, !# = gcd !, # ⋅ lcm(!, #)

lcm(2B ⋅ 3, 27⋅ 37) = 2B ⋅ 37 = 72

The Division Algorithm

Theorem: Let ! ∈ ℤ and $ ∈ ℤ%. Then there are unique &, ( ∈ ℤ, with 0 ≤ ( < $, such that

! = $& + (

Ex: ! = 101, $ = 2! = −11, $ = 3

& = ! div $( = ! mod $

divisor quotient remainder

= !/$= ! − $ !/$ $ | ! ⇔ ! mod $ = 0

The Division Algorithm

Theorem: Let ! ∈ ℤ and $ ∈ ℤ%. Then there are unique &, ( ∈ ℤ, with 0 ≤( < $, such that ! = $& + (1. Existence (5.2 Example 5): use the well-ordering property: “Every

nonempty subset of ℕ has a least element”

2. Uniqueness (exercise)

The Euclidean Algorithm

qA useful fact about the division algorithm:

Theorem: Let ! = #$ + &, where !, #, $, & ∈ ℤ. Then gcd !, # = gcd(#, &)

qA more efficient way to find gcd:

Euclidean Algorithm: find gcd !, # by successively applying the division algorithm

The Euclidean Algorithm

Ex: Find gcd 287,91 using the Euclidean Algorithm

287 = 91 ⋅ 3 + 1491 = 14 ⋅ 6 + 7⇒ gcd 287,91 = gcd(91,14) = gcd(14,7) = 7

gcd 287,91 = gcd(91,14)

gcd 91,14 = gcd(14,7)

GCDs as Linear CombinationsBezout’s Theorem: Let !, # ∈ ℤ&. There exist ', ( ∈ ℤ such that

gcd !, # = '! + (#

Ex: Find ', ( ∈ ℤ such that gcd 54,15 = ' ⋅ 54 + ( ⋅ 1554 = 3 ⋅ 15 + 915 = 1 ⋅ 9 + 69 = 1 ⋅ 6 + 3

9 = 54 − 3 ⋅ 156 = 15 − 1 ⋅ 93 = 9 − 1 ⋅ 6Backward substitution gives3 = 9 − 1 ⋅ 6= 9 − 1 ⋅ (15 − 1 ⋅ 9)= 2 ⋅ 9 − 1 ⋅ 15= 2 ⋅ 54 − 3 ⋅ 15 − 1 ⋅ 15= 2 ⋅ 54 − 7 ⋅ 15 ⇒ ' = 2, ( = −7

gcd 54,15 = gcd 15,9= gcd 9,6= gcd 6,3= 3

Applications of Bezout’s Theorem

Lemma: If !, #, $ ∈ ℤ' such that gcd !, # = 1 and ! | #$, then ! | $• We say that ! and # are relatively prime if gcd !, # = 1

Corollary: If . is a prime and . | !/!0 …!2 where each !3 is an integer, then . | !3for some 4.The Fundamental Theorem of Arithmetic: Every positive integer > 1 can be written uniquely as a prime or as the product of two or more primes where the primer factors are written in non-decreasing order

Proof: 1. existence: strong induction2. uniqueness: using the above corollary

Wrap Up1. Divisibility: ! | # ⇔ # = &! for some integer &2. Primes

• the Fundamental theorem of Arithmetic• A composite ' has a prime divisor ≤ '• there are infinite many primes

3. Greatest common divisor and least common multiple

4. Division algorithm: ! = )* + ,, 0 ≤ , < )• gcd !, ) = gcd(), ,)

5. Euclidean algorithm: find gcd by successively applying the division algorithm

6. Bezout’s Theorem: gcd !, # = 5! + 6#• If gcd !, # = 1 and ! | #8, then ! | 8

Congruences

Definition: Let !, # ∈ ℤ,& ∈ ℤ', we say ! is congruent to # modulo & if & | (! − #)• If ! is congruent to # modulo &,we write ! ≡ # (mod &)

• Examples

• 17 ≡ 5 mod 6 ?• 11 ≡ 8 mod 2 ?

• ! ≡ # mod & ⇔ & | (! − #)⇔ ! − # = :& for some : ∈ ℤ⇔ ! = :& + # for some : ∈ ℤ

14 ≡ 2 mod 1223 ≡ 11 (mod 12)

Congruences (cont.)

Theorem: Let !, #, $, % ∈ ℤ,( ∈ ℤ)

• ! ≡ # mod ( ⇔ (! mod () = (# mod ()• If ! ≡ # (mod () and # ≡ $ (mod (), then ! ≡ $ mod (

• If ! ≡ # (mod () and $ ≡ % (mod (), then ! + $ ≡ # + % (mod () and !$ ≡ #% (mod ()

Theorem: Let ! ∈ ℤ,( ∈ ℤ). There is a unique !3 ∈ {0,1, … ,( − 1} such that ! ≡ !3 (mod ().

Arithmetic Modulo !ℤ# = 0,1, … ,! − 1Addition modulo !: * +# , = * + , mod !Multiplication modulo !: * ⋅# , = * ⋅ , mod !

Ex: 6 +239, 7 ⋅22 8

• * +# , = 7 ⇒ * + , ≡ 7 mod !• * ⋅# , = 7 ⇒ * ⋅ , ≡ 7 (mod !)

Properties of ℤ"For any #, %, & ∈ ℤ"• Closure: # +" % ∈ ℤ"

# ⋅" % ∈ ℤ"

• Associativity: # +" % +" & = # +" (% +" &)# ⋅" % ⋅" & = # ⋅" (% ⋅" &)

• Commutativity: # +" % = % +" ## ⋅" % = % ⋅" #

Properties of ℤ"For any #, %, & ∈ ℤ"• Distributivity: # ⋅" % +" & = # ⋅" % +" # ⋅" &

(# +"%) ⋅" & = # ⋅" & +" % ⋅" &

• Identity elements: # +" 0 = 0 +" # = ## ⋅" 1 = 1 ⋅" # = #

• Additive inverse: For every # ∈ ℤ", there is % ∈ ℤ", such that # +" % = 00 +" 0 = 0# +" / − # = 0 for # ≠ 0

Properties of ℤ"• For # ∈ ℤ" , & ∈ ℤ" is a multiplicative inverse of # if # ⋅" & = 1,

• does 2 have a multiplicative inverse in ℤ+?

• does 2 have a multiplicative inverse modulo ℤ,?

• Theorem: # has a multiplicative inverse in ℤ" if and only if gcd #,0 = 1.

• Corollary: Every non-zero element has a multiplicative inverse in ℤ2 when 3 is prime

No

Yes 2 ⋅ 3 ≡ 1 mod 5

Additive Inverse and Multiplicative Inverse

• For $, & ∈ ℤ,• & is an additive inverse of $ modulo ) ∈ ℤ* if $ + & ≡ 0 mod )• & is an multiplicative inverse of $ modulo ) ∈ ℤ* if $ ⋅ & ≡ 1 mod )

• Theorem: $ ∈ ℤ and $ ≠ 0 has a multiplicative inverse modulo ) ∈ ℤ* if and only if gcd $,) = 1. Furthermore, an inverse, when it exists, is unique modulo ).

Find Multiplicative Inverses

Ex 1: Find a multiplicative inverse of 3 modulo 73# ≡ 1 ≡ 8 ≡ 15 (mod 7) ⇒ # ≡ 5 (mod 7)

Ex 2: Find a multiplicative inverse of 5 modulo 35# ≡ 1 ≡ 4 ≡ 7 ≡ 10 (mod 3) ⇒ # ≡ 2 mod 3

Use Bezout’s Theorem to find an inverse of 1 modulo 2, where gcd 1,2 = 1• find 7, 8 ∈ ℤ such that 71 + 82 = 1• 7 is a multiplicative inverse of 1 modulo 2

Ex 3: Find an inverse of 101 modulo 4620 (4.4 Example 2)

Solving Linear CongruencesProblem: Given !, # ∈ ℤ, & ∈ ℤ', find ( ∈ ℤ such that

!( ≡ # (mod &)

Let us first assume gcd !,& = 1.

Ex: Find the solution of 3( ≡ 4 mod 7

3( ≡ 4 ≡ 11 ≡ 18 mod 7⇒ ( ≡ 6 mod 7

We know 3 ⋅ 5 ≡ 1 mod 7Then 3( ≡ 4 mod 7⇒ 5 ⋅ 3( ≡ 5 ⋅ 4 (mod 7)⇒ ( ≡ 20 ≡ 6 (mod 7)

Solving Linear Congruences

Problem: Given !, # ∈ ℤ, & ∈ ℤ', find all ( ∈ ℤ such that

!( ≡ # (mod &)

Q: What if gcd !,& = 2 > 1?

A: For the linear congruence to have a solution, we must have 2 | #⇒We only need to solve !8( ≡ #8 mod &′ where !′ = :

; , #8 = <

; , and &8 = =;

Ex: Find the solution of 15( ≡ 6 mod 9

Modular Exponentiation and Fermat’s Little Theorem

Ex: Find 2" mod 7

Fermat’s Little Theorem: If ' is prime, then for every integer ( we have

() ≡ ( (mod ')Further, if ( is not divisible by ', then

()-. ≡ 1 (mod ')

ØSee 4.4 Exercise 19 for a proof sketch

Ex: Find 7000 mod 11

To compute (1 mod ' where ' is prime and ' ∤ (• First write 3 = 5 ' − 1 + 8 where 0 ≤ 8 < ' − 1• Then (1 = (< )-. =>

= (()-. )<(>≡ 1<(> (mod ')≡ (> (mod ')

Pierre de Fermat

Fast Modular Exponentiation

Ex: Find 3"# mod 645

36 = 2' + 2)

3)* mod 645 = 9

3)+ mod 645 = 9) mod 645 = 813)1 mod 645 = 81) mod 645 = 6561 mod 645 = 1113)5 mod 645 = 111) mod 645 = 12,321 mod 645 = 663)7 mod 645 = 66) mod 645 = 4356 mod 645 = 4863"# mod 645 = 3)7 ⋅ 3)+ mod 645 = 486 ⋅ 81 mod 645 = 21

Outline

• Divisibility and Modular Arithmetic (4.1)• Primes and GCD (4.3)• Solving Congruences (4.4)• Cryptography (4.6)

Introduction to Cryptography

• Classical Cryptography

• Shift Cipher • Affine Cipher

• Public Key Cryptography

• RSA

Symmetric Key Cryptography

Eve

Symmetric Key Cryptography

• Bob and Alice need to share the secret key !• Need to make sure " = $%('%("))

"

Type equationhere.7 = '% " " = $% 7

Bob

"Alice

7

Eve

encryption decryption

Shift Cipher

• Caesar Cipher: shift each letter three letters forward in the alphabet• Plain: ! " # $ % & …( ) * + , - .• Cipher: / 0 1 2 ℎ 4 …5 6 7 8 9 : ;• Ex: TULANE → 56=/>ℎ

• Mathematically, encode letters as numbers in ℤ@A = {0,1, … , 25}• ! " # $ % & … ) * + , - .• 0 1 2 3 4 5 … 20 21 22 23 24 25

• Encryption: ; = 0L M = M + O mod 26• Decryption: M = /L ; = ; − O mod 26• Do we have M = /L(0L(M))?

M: plaintext, ;: ciphertext, O: keyM, ;, O ∈ ℤ@A

Affine Cipher

• Encryption: ! = # ⋅ % + ' mod 26• #, ' is the key where #, ' ∈ ℤ01 and gcd #, 26 = 1

• Ex: # = 7, ' = 3, % = 10 (‘8’), what is !?

• Decryption: % = 9# ! − ' mod 26• 9# ∈ ℤ01, # 9# ≡ 1 (mod 26)

• Do we have % = >?(@?(%))?

! = 21 (‘v’)

Public Key Cryptography

Anyone can send a secret (encrypted) message to the receiver, without any prior contact, using publicly available info.

Albert R. Meyer March 13, 2013

Public Key Cryptography

• Invented by Diffie & Hellman in 1976

• They shared the 2015 Turing Award

• Why Public Key Cryptography?

• Key distribution

• Digital signature

Public Key Cryptography

• Alice has a key pair ! = !#$%, !#'() , Bob only knows !#$%• Need to make sure * = +,-./0(2,-34(*))

*

Type equation here.D = 2,-34 * * = +,-./0 D

Bob

*Alice

D

encryption decryption

Eve

The RSA Cryptosystem

• One of the first practical public key cryptosystems

• Invented by Ronald Rivest, Adi Shamir, and Lenoard Adleman in 1976

• They shared the 2002 Turing Award

• Based on the difficulty of factoring large numbers into primes

The RSA Cryptosystem

Message Encoding: 1. Each letter is encoded into a two-digit number

! " # … % & ' ( … ) * + , - . / 0 1 2 3 400 01 02 …08 09 10 11 …14 15 16 17 18 19 20 21 22 23 24 25

2. A message is divided into ? letter blocks such that the maximum 2? digits does not exceed @

Ex: @ = 2537, a message is divided into 2 letter blocks (2525 < 2537<252525)• Message STOP is translated into two blocks 1819 1415

Plain and cipher texts are numbers in ℤD = 0,1, … , @ − 1 .

The RSA Cryptosystem

Key generation (by Alice):1. Select two large primes !, #, ! ≠ #2. ' = ! ⋅ #3. Select a small odd integer * that is relatively prime to (! − 1)(# − 1)4. Compute / such that /* ≡ 1 (mod ! − 1 # − 1 )5. 5678 = ', * is the public key

6. 56:;< = (', /) is the private key

Ex: ! = 43 # = 59 ' = ! ⋅ # = 2537 * = 13 / = 3615678 = (2537, 13), 56:;< = (2537, 361)

RSA Encryption and DecryptionTo encrypt a plaintext ! use the public key (#, %)

' = !) mod #To decrypt a ciphertext ' use the private key (#, -)

! = '. mod #

Ex: Encrypt the message STOP with the public key (2537, 13)• Message STOP is translated into two blocks 1819 1415• Compute 181978 mod 2537, 141578 mod 2537 using fast modular exponentiation

Do we have ! = -9(%9(!))?

Security of RSA: It is hard to guess - given (#, %) (hard to factor # = :; for large : and ;)

Need to show !) . ≡ ! mod :; (Section 4.6)

Public Key Cryptography

• Alice has a key pair ! = !#$%, !#'() , Bob only knows !#$%• Need to make sure * = +,-./0(2,-34(*))

*

Type equation here.D = 2,-34 * * = +,-./0 D

Bob

*Alice

D

encryption decryption

Eve

Digital Signature

• Alice has a key pair ! = !#$%, !#'()• Need to make sure * = +,-./(1,-234(*))

*

Type equation here.* = +,-./ D D = 1,-234 *

Bob

*Alice

D

verification signing

Eve